Question

If $A+B=\frac{\pi }{4}$. Then $(1+\tan A)(1+\tan B)$$-(\cot A-1)(\cot B-1)+2$ equals

Answer 1

If $A+B=\frac{\pi }{4}$

taking $\tan$ both sides

$\tan (A+B)=\tan \frac{\pi }{4}=1$

$\Rightarrow \tan (A+B)=1$

$\Rightarrow \frac{\tan A+\tan B}{1-\tan A\tan B}=1$

$\Rightarrow \tan A+\tan B=1-\tan A\tan B$

$\Rightarrow \tan A+\tan B+\tan A\tan B+1=2$

$\Rightarrow \tan A+\tan A\tan B+1+\tan B=2$

$\Rightarrow \tan A(1+\tan B)+1(1+\tan B)=2$

$\Rightarrow (1+\tan A)(1+\tan B)=2⟶\left(1\right)$

Again,

$A+B=\frac{\pi }{4}$

taking $\cot$ of both sides

$\cot (A+B)=\cot \frac{\pi }{4}=1$

$\Rightarrow \frac{\cot A\cot B-1}{\cot B+\cot A}=1$

$\Rightarrow \cot A\cot B-1=\cot A+\cot B$

$\Rightarrow \cot A\cot B-\cot B-\cot A+1=1+1$

$\Rightarrow \cot B(\cot A-1)-1(\cot A-1)=2$

$\Rightarrow (\cot A-1)(\cot B-1)=2⟶\left(2\right)$

Now,

$(1+\tan A)(1+\tan B)-(\cot A-1)(\cot B-1)+2$

putting values from $\left(1\right)$ and $\left(2\right)$

$\Rightarrow 2-2+2$

$=2$